Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If `(C_A and C_B) be the molar heat capacities for the two processes,

Options

A. `C_A = C_B`
B. `C_A lt C_B`
C. `C_A gtC_B`
D. `C_A and C_B` cannot be defined.

Answer

To solve the problem, we will analyze the relationship between the work done by an ideal gas in two different processes and the corresponding molar heat capacities. Here are the steps to arrive at the solution:

### Step 1: Define the Work Done in Both Processes
Let the work done in process A be \( W_A \) and in process B be \( W_B \). According to the problem, we have:
\[ W_A = 2 W_B \]

### Step 2: Relate the Change in Internal Energy
Since the temperature rises by the same amount in both processes, we can denote the change in temperature as \( \Delta T \). The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = n C_V \Delta T \]
where \( C_V \) is the molar heat capacity at constant volume and \( n \) is the number of moles. Since the temperature change is the same in both processes, we have:
\[ \Delta U_A = \Delta U_B \]

### Step 3: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
For processes A and B, we can write:
\[ Q_A = \Delta U_A + W_A \]
\[ Q_B = \Delta U_B + W_B \]

### Step 4: Express Heat Added in Terms of Heat Capacities
The heat added in each process can also be expressed in terms of the molar heat capacities:
\[ Q_A = n C_A \Delta T \]
\[ Q_B = n C_B \Delta T \]

### Step 5: Substitute and Rearrange
Substituting the expressions for \( Q \) into the first law equations gives:
\[ n C_A \Delta T = n C_V \Delta T + W_A \]
\[ n C_B \Delta T = n C_V \Delta T + W_B \]

### Step 6: Isolate Work Done
Rearranging these equations, we find:
\[ W_A = n C_A \Delta T - n C_V \Delta T \]
\[ W_B = n C_B \Delta T - n C_V \Delta T \]

### Step 7: Substitute Work Done Relationship
Now, substituting \( W_A = 2 W_B \) into the first equation:
\[ n C_A \Delta T - n C_V \Delta T = 2(n C_B \Delta T - n C_V \Delta T) \]

### Step 8: Simplify the Equation
This simplifies to:
\[ n C_A \Delta T - n C_V \Delta T = 2n C_B \Delta T - 2n C_V \Delta T \]
\[ n C_A \Delta T + n C_V \Delta T = 2n C_B \Delta T \]

### Step 9: Factor Out Common Terms
Dividing through by \( n \Delta T \) (assuming \( n \) and \( \Delta T \) are not zero):
\[ C_A + C_V = 2 C_B \]

### Step 10: Rearranging for Comparison
Rearranging gives:
\[ C_A = 2 C_B - C_V \]

### Conclusion
Since \( C_V \) is always positive, we conclude that:
\[ C_A > C_B \]

Thus, the molar heat capacity for process A is greater than that for process B.